Question: Divide the following complex numbers. $ \dfrac{7-24i}{-3-4i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3+4i}$ $ \dfrac{7-24i}{-3-4i} = \dfrac{7-24i}{-3-4i} \cdot \dfrac{{-3+4i}}{{-3+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(7-24i) \cdot (-3+4i)} {(-3-4i) \cdot (-3+4i)} = \dfrac{(7-24i) \cdot (-3+4i)} {(-3)^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(7-24i) \cdot (-3+4i)} {(-3)^2 - (-4i)^2} = $ $ \dfrac{(7-24i) \cdot (-3+4i)} {9 + 16} = $ $ \dfrac{(7-24i) \cdot (-3+4i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({7-24i}) \cdot ({-3+4i})} {25} = $ $ \dfrac{{7} \cdot {(-3)} + {-24} \cdot {(-3) i} + {7} \cdot {4 i} + {-24} \cdot {4 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{-21 + 72i + 28i - 96 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{-21 + 72i + 28i + 96} {25} = \dfrac{75 + 100i} {25} = 3+4i $